Question

In Exercises 5–16, test the given claim.

Body Temperatures of Men and Women If we use the body temperatures from 8 AMon Day 2 as listed in Data Set 3 “Body Temperatures” in Appendix B, we get the statistics given in the accompanying table. Use these data with a 0.05 significance level to test the claim that men have body temperatures that vary more than the body temperatures of women.

Male	Female
\({n_1} = 11\)	\({n_2} = 59\)
\({\bar x_1} = {97.69^ \circ }F\)	\({\bar x_2} = {97.45^ \circ }F\)
\({s_1} = {0.89^ \circ }F\)	\({s_1} = {0.66^ \circ }F\)

Short answer

There is not enough evidence to support the claim that men have body temperatures that vary more than the body temperatures of women.

Step-by-step solution

Given information

For a sample of 11 body temperatures of men, the mean temperature is equal to97.69 degrees Fahrenheit,and the standard deviation of temperatures is equal to 0.89 degrees Fahrenheit.For another sample of 59 body temperatures of women, the mean temperature is equal to 97.45 degrees Fahrenheitand the standard deviation of temperatures is equal to 0.66 degrees Fahrenheit.

It is claimed that the variation in men’s body temperature is more than the variation in women’s body temperature.

Hypotheses

Let\({\sigma _1}\)and\({\sigma _2}\)be the population standard deviationsof the men’s body temperaturesand the women’s body temperature,respectively.

Null Hypothesis: The population standard deviation of men’s body temperature is equal to the population standard deviation of women’s body temperature.

Symbolically,

\({H_0}:{\sigma _1} = {\sigma _2}\)

Alternate Hypothesis: The population standard deviation of men’s body temperature is greater than the population standard deviation of women’s body temperature.

Symbolically,

\({H_1}:{\sigma _1} > {\sigma _2}\)

Compute the test statistic

Since two independent samples involve a claim about the population standard deviation, apply an F-test.

Consider the larger sample variance to be\(s_1^2\)and the corresponding sample size to be\({n_1}\).

The following values are obtained:

\({\left( {0.89} \right)^2} = 0.7921\)

\({\left( {0.66} \right)^2} = 0.4356\)

Here\(s_1^2\)is the sample variance corresponding to men’s body temperature and has a value equal to 0.7921\({\left( {^ \circ F} \right)^2}\).

\(s_2^2\)is the sample variance corresponding to women’s body temperature and has a value equal to 0.4356\({\left( {^ \circ F} \right)^2}\).

Substitute the respective values to calculate the F statistic:

\(\begin{array}{c}F = \frac{{s_1^2}}{{s_2^2}}\\ = \frac{{{{\left( {0.89} \right)}^2}}}{{{{\left( {0.66} \right)}^2}}}\\ = 1.818\end{array}\)

State the critical value and the p-value

The value of the numerator degrees of freedom is equal to:

\(\begin{array}{c}{n_1} - 1 = 11 - 1\\ = 10\end{array}\)

The value of the denominator degrees of freedom is equal to:

\(\begin{array}{c}{n_2} - 1 = 59 - 1\\ = 58\end{array}\)

For the F test, the critical value corresponding to the right-tail is considered.

The critical value can be obtained using the F-distribution table with numerator degrees of freedom equal to 10and denominator degrees of freedom equal to 58 for a right-tailed test.

The level of significance is equal to 0.05.

Thus, the critical value is equal to 1.9983.

The right-tailed p-value for F equal to 1.818 is equal to 0.0775.

Conclusion

Since the test statistic value is less than the critical value and the p-value is greater than 0.05, the null hypothesis is failed to reject.

Thus, there is not enough evidence to supportthe claimthat men have body temperatures that vary more than the body temperatures of women.